Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $r = \dfrac{z + 1}{z + 9} \times \dfrac{-z^2 - 14z - 45}{-z^2 + 2z + 3} $
First factor out any common factors. $r = \dfrac{z + 1}{z + 9} \times \dfrac{-(z^2 + 14z + 45)}{-(z^2 - 2z - 3)} $ Then factor the quadratic expressions. $r = \dfrac {z + 1} {z + 9} \times \dfrac {-(z + 9)(z + 5)} {-(z + 1)(z - 3)} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac {(z + 1) \times -(z + 9)(z + 5) } {(z + 9) \times -(z + 1)(z - 3) } $ $r = \dfrac {-(z + 9)(z + 5)(z + 1)} {-(z + 1)(z - 3)(z + 9)} $ Notice that $(z + 1)$ and $(z + 9)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac {-(z + 9)(z + 5)\cancel{(z + 1)}} {-\cancel{(z + 1)}(z - 3)(z + 9)} $ We are dividing by $z + 1$ , so $z + 1 \neq 0$ Therefore, $z \neq -1$ $r = \dfrac {-\cancel{(z + 9)}(z + 5)\cancel{(z + 1)}} {-\cancel{(z + 1)}(z - 3)\cancel{(z + 9)}} $ We are dividing by $z + 9$ , so $z + 9 \neq 0$ Therefore, $z \neq -9$ $r = \dfrac {-(z + 5)} {-(z - 3)} $ $ r = \dfrac{z + 5}{z - 3}; z \neq -1; z \neq -9 $